Let $f(x)=\dfrac{\sqrt x}{\sin(x)}$. Find $f'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\sin(x)+2x\cos(x)}{2\sqrt x\sin^2(x)}$ (Choice B) B $\dfrac{\cos(x)}{2\sqrt x}$ (Choice C) C $\dfrac{\sin(x)-2x\cos(x)}{2\sqrt x\sin^2(x)}$ (Choice D) D $\dfrac{1}{2\sqrt x\cos(x)}$
Explanation: $f(x)$ is the quotient of two, more basic, expressions: $\sqrt x$ and $\sin(x)$. Therefore, the derivative of $f$ can be found using the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! $\begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}\left[\dfrac{\sqrt x}{\sin(x)}\right] \\\\ &=\dfrac{\dfrac{d}{dx}[\sqrt x]\sin(x)-\sqrt x\dfrac{d}{dx}[\sin(x)]}{[\sin(x)]^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{\dfrac{1}{2\sqrt x}\cdot\sin(x)-\sqrt x\cdot\cos(x)}{\sin^2(x)}&&\gray{\text{Differentiate }\sqrt x\text{ and }\sin(x)} \\\\ &=\dfrac{\sin(x)-2x\cos(x)}{2\sqrt x\sin^2(x)}&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $f'(x)=\dfrac{\sin(x)-2x\cos(x)}{2\sqrt x\sin^2(x)}$